Jordan form, number of blocks. Suppose I have an eigenvalue $\lambda$, now I want to determine the number of Jordan blocks corresponding to that eigenvalue, as well as size of each block. I know that:

*

*number of blocks is equal to the number of eigenvectors

*there are $\dim \ker (A - \lambda I)^k - \dim \ker (A - \lambda I)^{k-1}$ blocks of size at least $k$
Why exactly are those statements true. Could someone provide me with a proof or at least some reasoning behind this?

 A: To answer your first question, each block looks like
$$ \left( \begin{array}{ccccc} 
\lambda & 1 & & & \\
 & \lambda & 1 & & \\
 & & \ddots & \ddots & \\
 & & & \lambda & 1 \\
 & & & & \lambda
\end{array} \right) $$
The columns of a matrix correspond to the basis vectors. You can see here that the first column gives an eigenvector, and the rest do not. The columns here tell you that
$$ \begin {align*}
Ae_1 &= \lambda e_1 \\
Ae_2 &= e_1 + \lambda e_2 \\
     &\vdots \\
Ae_n &= e_{n-1} + \lambda e_n
\end {align*} $$
So each Jordan block gives you exactly 1 eigenvector.
A: To answer the second question, one can assume that $\lambda =0$ (since we are always considering $A-\lambda I$). Let $J$ be the usual Jordan block of size $n$ and eigenvalue zero and note that $\dim\ker J^k = k$. To see this, note that $J^k$ consists of a diagonal of ones of length $n-k$. In particular, it follows that
$$\dim \ker J^k - \dim \ker J^{k-1} =\begin{cases} 1 & \text{ if $1\leqslant k\leqslant n$ } \\
 0 & \text{ if $k>n$. }
\end{cases}
$$
Hence, if you let $b_k(J) = \dim \ker J^k - \dim \ker J^{k-1}$, this is counting the number of blocks of size at least $k$ in $J$, and it follows that $b_k(J) - b_{k+1}(J)$ counts the number of blocks of size exactly $k$ in $J$: by the formula above it is zero except when $k=n$.
Since conjugation preserves $b_k(A) = \dim \ker A^k$, and since dimensions add up when putting matrices in blocks, it follows that $b_k(A-\lambda)-b_{k+1}(A-\lambda)$ counts the number of blocks of size at least $n$ in $A$ corresponding to the eigenvalue $\lambda$.
(Note that we do not need to care about other eigenvalues in the formula since $J(\lambda)^k$ is always full rank unless $\lambda=0$.)
